Question

A radioactive nucleus undergoes $\alpha$-emission to form a stable element. What will be the recoil velocity of the daughter nucleus if $v$ is the velocity of $\alpha$ emission?

• A. $\frac{4v}{A-4}$
• B. $\frac{2v}{A-4}$
• C. $\frac{4v}{A+4}$
• D. $\frac{2v}{A+4}$

Answer 1

The correct option is A $\frac{4v}{A-4}$

We assume that mass number of nucleus when it was at rest = $A$
$\because$mass number of $\alpha$ =particle = 4
$\therefore$mass number of remaining nucleus = $A$ - 4
As there is no external force, so momentum of the system will remain conserved
$\Rightarrow$$0=(A-4)v^{{}^{\prime }}+4v\Rightarrow v^{{}^{\prime }}=-\frac{4v}{(A-4)}$
negative sign represents that direction is opposite to the direction of motion of $\alpha$ -particle.